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Algebra and absolute value question I'm stuck on

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by beboppin » Sat Aug 15, 2009 10:30 am
What is the product of all the solutions of (x^2) - 4x + 6 = 3 - |x-1| ?

This is a question from Total GMAT. The answer I've come up with is not an available solution. I see the explanation in the back but I think I'm missing something here. Any help would be much appreciated.
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by scoobydooby » Sat Aug 15, 2009 11:15 am
is it 8?

(x^2) - 4x + 6 = 3 - |x-1|
=>(x^2) - 4x + 6-3=-|x-1|
=>x^2 -4x+3=-|x-1|
=>(x-1)(x-3)=-|x-1|
squaring both sides

(x-1)^2 (x-3)^2=(x-1)^2
=>(x-1)^2*[(x-3)^2-1]=0
=>(x-1)^2*[x^2+9-6x-1]=0
=>(x-1)^2*(x-4)(x-2)=0
=>x=1 or x=4 or x=2

product of all possible solutions: 1*4*2=8
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by beboppin » Sat Aug 15, 2009 12:35 pm
scoobydooby wrote:is it 8?

(x^2) - 4x + 6 = 3 - |x-1|
=>(x^2) - 4x + 6-3=-|x-1|
=>x^2 -4x+3=-|x-1|
=>(x-1)(x-3)=-|x-1|
squaring both sides

(x-1)^2 (x-3)^2=(x-1)^2
=>(x-1)^2*[(x-3)^2-1]=0
=>(x-1)^2*[x^2+9-6x-1]=0
=>(x-1)^2*(x-4)(x-2)=0
=>x=1 or x=4 or x=2

product of all possible solutions: 1*4*2=8
Hmm..that is what the book said too. But if you plug 4 in, the equation does not work, right? 6 = 0?
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by scoobydooby » Sat Aug 15, 2009 9:11 pm
yes you are right. 4 doesnt work when put back in the original equation.
1 and 2 are the only possible solutions.

which book was the question from?
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by beboppin » Sun Aug 16, 2009 6:36 am
scoobydooby wrote:yes you are right. 4 doesnt work when put back in the original equation.
1 and 2 are the only possible solutions.

which book was the question from?
Total GMAT Math. I do want to caveat that this book has been great otherwise.

So is the method of assuming an absolute value term is positive and then assuming it's negative something I should avoid looking to do? I'd never gone down that route before and am wondering why it doesn't work in this case, assuming that it doesn't.
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by tohellandback » Sun Aug 16, 2009 9:02 am
scoobydooby wrote:is it 8?

(x^2) - 4x + 6 = 3 - |x-1|
=>(x^2) - 4x + 6-3=-|x-1|
=>x^2 -4x+3=-|x-1|
=>(x-1)(x-3)=-|x-1|
squaring both sides

(x-1)^2 (x-3)^2=(x-1)^2
=>(x-1)^2*[(x-3)^2-1]=0
=>(x-1)^2*[x^2+9-6x-1]=0
=>(x-1)^2*(x-4)(x-2)=0
=>x=1 or x=4 or x=2

product of all possible solutions: 1*4*2=8
sccobydooby,
you gotta eliminate the answer here in this line:
(x-1)(x-3)=-|x-1|
|x-1| is always>=0
so -|x-1|<=0
so (x-1)(x-3)<=0
so 1<=x<=3 (for any values "inside" the roots on the number lines, the result is negative for any quadratic equation)
from the answer choices you have got finally, only x=1 and 2 satisfy
answer should be 2
The powers of two are bloody impolite!!
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